Solution - dm2_080303

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Calculate the effective width of plate from AISC Manual
of Steel Construction, Appendix C, Equation C3-1.
be = (253t)/(fy1/2)[1 - 50.3/(b/t)(fy1/2)]
= 253(3/8)/361/2
[1 - 50.3/(16/(3/8)(361/2)]
= 12.70 in
(14) Calculate the elastic and plastic section moduli of the
combined section.
_
Let y be the distance of c.g. of the combined section from
the outside edge of the plate as shown in Figure 77.
_
y = (12.70 x 3/8 x 3/16) + (6 + 3/8 - 3) x 2.40
(12.70 x 3/8) + 2.40
= 1.256 inches
Let yp be the distance to the N.A. of the combined
section for full plasticity.
12.70 yp = 2.40 + 12.70(3/8 - yp)
yp = 7.163/25.4
= 0.282 in.
I = [12.70 x (3/8)3]/12 + 12.70(3/8)x (1.256 -
3/16)2 + 13.1 + 2.40 x (3/8 + 3 - 1.256)2
= 29.37 in4
Hence,
Smin = 29.37/(6.375 - 1.256)
= 5.737 in3
Z = 12.70(0.282)2/2 + 12.70(3/8 - 0.282)2/2
+ 2.40(6.375 - 3 - 0.282)
= 7.983 in3
(15) Mp = 39.6(5.737 + 7.983)/2
= 271.6 in-k
Calculate the ultimate dynamic shear capacity, Vp.
Vp = FdvAw
= 21.78(4.375 x 0.20)
= 19.06 k
Calculate support shear and check shear capacity.
L = 3 ft or 36 in
V = (27.0 x 16.0 x 36)/2
= 7,776 lb
7.776 k < Vp
O.K.
2.08-286